When do I differentiate and when do I integrate in AP Physics C?

Differentiate to go down the motion chain (position to velocity to acceleration) or to get a force from a potential energy ($F = -dU/dx$). Integrate to go up the chain (acceleration to velocity to position), to find work from a variable force ($\int F\,dx$) or impulse from a force in time ($\int F\,dt$), and to find the center of mass or moment of inertia ($\int x\,dm$, $\int r^2\,dm$). A given function of time means differentiate; a rate or density to accumulate means integrate.

How do I set up a moment-of-inertia or center-of-mass integral?

Write the mass element $dm$ in terms of the geometry and density: $dm = \lambda\,dx$ for a rod, $\sigma\,dA$ for a sheet, $\rho\,dV$ for a solid. Then integrate the appropriate quantity over the body: $x_{cm} = \tfrac{1}{M}\int x\,dm$ for the center of mass, $I = \int r^2\,dm$ for the moment of inertia, with $r$ the distance from the axis. Choose limits that cover the whole body and simplify using the uniform density where possible.

How do I handle a force that depends on velocity, like air resistance?

Write Newton's second law as a differential equation, $m\,dv/dt = \sum F$, with the velocity-dependent term included (for example $mg - bv$ for a falling object with linear drag). Find the terminal velocity by setting the acceleration to zero, then, if asked, separate variables and integrate to get $v(t)$, which for linear drag is $v_t(1 - e^{-bt/m})$. The key is recognizing that the acceleration is not constant, so kinematic equations do not apply.

What calculus do I need for simple harmonic motion?

Recognize the defining differential equation $\ddot{x} = -\omega^2 x$, which comes from a linear restoring force. Confirm that $x = A\cos(\omega t + \phi)$ solves it by differentiating twice, and read off $\omega$. Differentiate the position to get velocity ($v = -A\omega\sin$) and acceleration ($a = -A\omega^2\cos$), which give $v_{max} = A\omega$ and $a_{max} = A\omega^2$. For a pendulum, derive the equation from the rotational second law and apply the small-angle approximation $\sin\theta \approx \theta$.

How are derivation questions marked on AP free-response?

AP graders award points for the reasoning, not just the final expression. Start from a stated principle (Newton's second law, conservation of energy), show each calculus step (the integral set up with limits, the derivative taken), keep the work symbolic until the end, and substitute numbers only in the final step. A clear derivation a marker can follow earns partial credit even with an algebra slip, so always show the physics and the calculus, not just the answer.

United StatesPhysics C: Mechanics

AP Physics C Mechanics using calculus in mechanics problems: a complete skills guide to derivatives, integrals and differential equations across the course

A deep-dive AP Physics C: Mechanics skills guide to the calculus that runs through the whole course: differentiating position for velocity and acceleration, integrating acceleration and variable forces, the force-potential relationship, moment-of-inertia and center-of-mass integrals, and differential equations for resistive forces and oscillations, with worked examples and exam technique.

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Why calculus is the spine of AP Physics C
Differentiate to go down the motion chain
Integrate to go up the chain (and to accumulate)
The force-potential relationship
Mass-element integrals
Differential equations: resistive forces and oscillations
Check your knowledge

Why calculus is the spine of AP Physics C

AP Physics C: Mechanics is the calculus-based course, and calculus is not an add-on: it is the language the whole course is written in. Velocity and acceleration are derivatives; work, impulse, the center of mass and the moment of inertia are integrals; resistive forces and oscillations are differential equations. Almost every distinctive AP Physics C exam question, the ones that separate it from the algebra-based AP Physics 1, turns on a calculus step. This guide ties together the calculus moves that recur across the units, each of which has its own dot-point page with practice: displacement, velocity and acceleration, work, potential energy, resistive forces, rotational inertia, and defining simple harmonic motion.

Differentiate to go down the motion chain

The first calculus skill is differentiation along the kinematic chain. Position differentiates to velocity, which differentiates to acceleration:

v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2}.

If a problem gives you the position as a function of time, you find velocity and acceleration by differentiating, usually polynomials, sometimes sines, cosines or exponentials. The same move gives a force from a potential energy: $F = -dU/dx$ . Whenever you are handed a function and asked for its rate of change, you differentiate.

Integrate to go up the chain (and to accumulate)

The reverse skill is integration, with the initial conditions supplying the constants:

v(t) = v_0 + \int_0^t a\,dt', \qquad x(t) = x_0 + \int_0^t v\,dt'.

Integration also accumulates a quantity over a body or an interval. Work is the integral of a variable force over distance, $W = \int F\,dx$ ; impulse is the integral of force over time, $J = \int F\,dt$ ; the center of mass and moment of inertia are integrals over mass elements, $x_{cm} = \tfrac{1}{M}\int x\,dm$ and $I = \int r^2\,dm$ . The recurring decision is simple: a function of time to differentiate means differentiate; a rate or density to add up means integrate.

The force-potential relationship

Conservative forces and potential energy are linked by calculus in both directions:

F(x) = -\frac{dU}{dx}, \qquad U(x) = -\int F(x)\,dx.

Given $U(x)$ , differentiate to find the force and set $F = 0$ to locate equilibria; the sign of $d^2U/dx^2$ tells you whether each is stable (minimum) or unstable (maximum). Given $F(x)$ , integrate to build the potential energy. This two-way relationship appears in energy problems and in the analysis of potential-energy curves.

Mass-element integrals

Two integrals over a body are signature AP Physics C tasks. For the center of mass and the moment of inertia:

x_{cm} = \frac{1}{M}\int x\,dm, \qquad I = \int r^2\,dm.

The trick is writing $dm$ through the density and geometry: $dm = \lambda\,dx$ (rod), $\sigma\,dA$ (sheet), $\rho\,dV$ (solid). Set the limits to span the body, and simplify. For a uniform rod about its center this gives $I = \tfrac{1}{12}ML^2$ ; about an end, $\tfrac{1}{3}ML^2$ . The parallel-axis theorem $I = I_{cm} + Md^2$ then shifts the axis without re-integrating.

Differential equations: resistive forces and oscillations

The deepest calculus in the course is setting up and solving differential equations. Two cases dominate.

Resistive forces. For a falling object with linear drag, Newton's second law is $m\,dv/dt = mg - bv$ . Setting the acceleration to zero gives the terminal velocity $v_t = mg/b$ ; solving the equation gives $v(t) = v_t(1 - e^{-bt/m})$ , an exponential approach.
Simple harmonic motion. A linear restoring force gives $\ddot{x} = -\omega^2 x$ , whose solution is $x = A\cos(\omega t + \phi)$ . You confirm the solution by differentiating twice and read off $\omega = \sqrt{k/m}$ (spring) or, for a pendulum, derive the equation from the rotational second law and apply $\sin\theta \approx \theta$ .

In both cases the essential recognition is that the acceleration is not constant, so the kinematic equations fail and you must work with the differential equation directly.

A variable force and the work-energy theorem

A $2.0$ kg particle starts from rest at $x = 0$ and is pushed along a frictionless track by a force $F(x) = (10 - 2.0x)$ N from $x = 0$ to $x = 5.0$ m. Find its speed at $x = 5.0$ m.

step 1 Recognize the calculus move

The force depends on position, so the work is the integral $W = \int F\,dx$ , not $Fd$ .

step 2 Set up and evaluate the work integral

$W = \int_0^{5.0}(10 - 2.0x)\,dx = \left[10x - x^2\right]_0^{5.0} = 50 - 25 = 25$ J.

step 3 Apply the work-energy theorem

The particle starts from rest, so $W = \tfrac{1}{2}mv^2 - 0$ : $25 = \tfrac{1}{2}(2.0)v^2 = v^2$ .

step 4 Solve for the speed

$v = \sqrt{25} = 5.0$ m/s. (Note the force is positive up to $x = 5.0$ m, where it just reaches zero, so the particle is still speeding up throughout.)

Setting up a resistive-force differential equation

An object of mass $0.50$ kg falls from rest through air with linear drag $b = 0.10$ kg/s. Take $g = 9.8$ m/s squared. Find the terminal velocity and write the velocity as a function of time.

step 1 Write Newton's second law as a differential equation

Down positive: $m\dfrac{dv}{dt} = mg - bv$ , i.e. $\dfrac{dv}{dt} = g - \dfrac{b}{m}v$ .

step 2 Find the terminal velocity

Set the acceleration to zero: $v_t = \dfrac{mg}{b} = \dfrac{(0.50)(9.8)}{0.10} = 49$ m/s.

step 3 Write the solution

The standard solution with $v(0) = 0$ is $v(t) = v_t\left(1 - e^{-bt/m}\right) = 49\left(1 - e^{-0.20t}\right)$ m/s.

step 4 Check the limits

At $t = 0$ , $v = 0$ (correct start); as $t \to \infty$ , $v \to 49$ m/s (terminal velocity). The time constant is $m/b = 5.0$ s.

Calculus is the spine of AP Physics C: Mechanics. Differentiate to go down the motion chain (position to velocity to acceleration) and to get force from potential energy ( $F = -dU/dx$ ). Integrate to go up the chain (with initial conditions) and to accumulate: work $\int F\,dx$ , impulse $\int F\,dt$ , center of mass $\tfrac{1}{M}\int x\,dm$ , moment of inertia $\int r^2\,dm$ (writing $dm$ through the density). Set up differential equations for velocity-dependent forces ( $m\,dv/dt = mg - bv$ , terminal velocity $mg/b$ , solution $v_t(1 - e^{-bt/m})$ ) and for simple harmonic motion ( $\ddot{x} = -\omega^2 x$ , solution $A\cos(\omega t + \phi)$ ). On derivations, start from a principle, keep it symbolic, and substitute only at the end.

Check your knowledge

A mix of calculus-based recall and calculation questions spanning the course. Attempt them under timed conditions, then check against the solutions.

State whether you differentiate or integrate to get velocity from acceleration, and what extra information you need. (2 marks)
A particle has $x(t) = 3t^2 - t^3$ (SI units). Calculate its acceleration at $t = 1.0$ s. (2 marks)
A force $F(x) = 4x$ N acts from $x = 0$ to $x = 3.0$ m. Calculate the work done. (2 marks)
State the relationship between a conservative force and its potential energy. (1 mark)
A uniform rod of mass $M$ and length $L$ has $I = \tfrac{1}{3}ML^2$ about one end. State the integral used to derive this. (2 marks)
An object falling with linear drag has terminal velocity when which condition holds? (1 mark)
Write the differential equation that defines simple harmonic motion. (1 mark)
An oscillator has $x = 0.05\cos(10t)$ (SI units). Calculate its maximum speed. (2 marks)
State the constant of integration's physical meaning when integrating acceleration to find velocity. (1 mark)
A solid sphere has $I = \tfrac{2}{5}MR^2$ . Explain why this is smaller than a hoop's $MR^2$ in terms of the integral $\int r^2\,dm$ . (2 marks)

Solutions

Q1: Integrate acceleration with respect to time; you need the initial velocity $v_0$ as the constant of integration: $v = v_0 + \int a\,dt$ .
Q2: $v = 6t - 3t^2$ , $a = 6 - 6t$ ; at $t = 1.0$ s, $a = 6 - 6 = 0$ m/s squared.
Q3: $W = \int_0^{3.0} 4x\,dx = [2x^2]_0^{3.0} = 2(9.0) = 18$ J.
Q4: The force is the negative derivative of the potential energy: $F = -dU/dx$ (equivalently $U = -\int F\,dx$ ).
Q5: $I = \int_0^L x^2\,dm$ with $dm = \lambda\,dx = \tfrac{M}{L}\,dx$ , giving $\tfrac{M}{L}\int_0^L x^2\,dx = \tfrac{M}{L}\cdot\tfrac{L^3}{3} = \tfrac{1}{3}ML^2$ .
Q6: When the net force is zero, that is when the drag equals the weight ( $bv = mg$ ), so the acceleration is zero.
Q7: $\dfrac{d^2x}{dt^2} = -\omega^2 x$ (the acceleration is proportional to the negative of the displacement).
Q8: $v_{max} = A\omega = (0.05)(10) = 0.50$ m/s.
Q9: It is the initial velocity $v_0$ , the velocity at the start of the interval, which the indefinite integral alone cannot determine.
Q10: $I = \int r^2\,dm$ weights mass by $r^2$ . The sphere has much of its mass at small $r$ (near the center), so the $r^2$ -weighted sum is smaller than the hoop's, whose mass all sits at $r = R$ .

Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)